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I come, once again looking for a tutorial. It seems like my discrete math class is lacking in explanations for more complex (to me) problems. My question is:

Compute the value of the following alternating sum and show all your work.

$$\binom{30}{3}4-\binom{30}{4}16+\binom{30}{5}64\mp \cdots-\binom{30}{30}4^{28}$$

Leave expressions of the form $m^n$ in your answer without attempting to evaluate them. Solutions that rely solely on calculator computation of the terms of the sum are unacceptable.

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Please do not answer the exact question, I would like to find the solution on my own. If someone could only show how to work this problem, I would be grateful. Thank you.

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  • $\begingroup$ Think about what $a(1+x)^n$ will look like for different $a,x,n$. $\endgroup$ – TokenToucan Feb 20 '16 at 1:49
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\begin{align} & \binom{30}{3}4-\binom{30}{4}16+\binom{30}{5}64\mp \cdots-\binom{30}{30}4^{28} \\[10pt] = {} & \frac {-1} {16} \left( \binom{30}{3}(-4)^3+\binom{30}{4}(-4)^4+\binom{30}{5} (-4)^5 + \cdots+\binom{30}{30}(-4)^{30} \right)\\[10pt] = {} & \frac {-1} {16} \left( \underbrace{\binom{30}{0}(-4)^0+\binom{30} 1 (-4)^1 + \binom{30} 2 (-4)^2 + \cdots+\binom{30}{30}(-4)^{30}} - \, \text{(three terms)} \right) \end{align} The binomial theorem handles the part over the $\underbrace{\text{underbrace}}$.

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Think of the binomial theorem...

$$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$

clearly there is an alternating sum, so we can let $y=-1$. That yields

$$(x-1)^n=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}x^k$$

Now, consider what $n$ is and consider what $x$ is... you may have to subtract terms off the summation, and consider a scalar $b$...

$$b(x-1)^{30}=\sum_{k=0}^{30}\binom{30}{k}(-1)^{30-k}x^k$$

Note that also when $k$ is odd, we have positive terms and when $k$ is even we have negative terms...that should set off bells of what $b$ should be...

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